
Two wires of different materials each of length l and cross sectional area ‘A’ are joined in series to form a composite wire. If their Young’s modulus are Y and 2Y, the total elongation produced by applying a force F to stretch the composite wire.
Answer
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Hint: The formula for young’s modulus has been used. The elongation for both the wire should be calculated. Then, the total elongation should be calculated. Substituting the values in the formula we will get the total elongation.
Formula used:
\[Y=\dfrac{Fl}{A\Delta l}\]
Complete answer:
We have two wires. These two wires are attached with each other. Both these wires are made up of different materials. So this is called composite wire.
The length of each wire is \[l\] and their area of cross section is \[A\] . The wires are connected in series. The young’s modulus of wire is given by Y and 2Y. Now applying force \[F\]across the wire. Due to the force applied, the wire will be stretched. So we need to calculate the elongation produced on wire due to the force applied.
Let us consider elongation in first wire of young’s modulus Y is \[\Delta {{l}_{1}}\] and that of 2Y is \[\Delta {{l}_{2}}\]
Let the total elongation be \[\Delta l\]
Therefore \[\Delta l=\Delta {{l}_{1}}+\Delta {{l}_{2}}\]….(1)
We know young’ s modulus is ratio of stress to strain
\[Y=\dfrac{{}^{F}/{}_{A}}{{}^{\Delta l}/{}_{l}}=\dfrac{Fl}{A\Delta l}\]
Substituting the values from equation (1)
\[\Delta l=\dfrac{Fl}{AY}\]
The force across both the wires will be the same. Therefore, tension in both the wires will be the same because it is a composite.
For first wire \[\Delta {{l}_{1}}=\dfrac{Fl}{AY}\]
Similarly for second wire \[\Delta {{l}_{2}}=\dfrac{Fl}{A(2Y)}\]
Substituting value in equation (1)
\[\begin{align}
& \Delta l=\dfrac{Fl}{AY}+\dfrac{Fl}{A(2Y)} \\
& \Delta l=\dfrac{3}{2}\dfrac{Fl}{AY} \\
\end{align}\]
Thus on applying force, the total elongation on wire will be \[\dfrac{3}{2}\dfrac{Fl}{AY}\].
Hence option D is correct.
Note:
Young’s modulus describes the physical change in a solid. It is a mechanical property of any solid. Young’s modulus depends on the force applied on the solid and the change in its length. Young’s modulus is different for different materials.
Formula used:
\[Y=\dfrac{Fl}{A\Delta l}\]
Complete answer:
We have two wires. These two wires are attached with each other. Both these wires are made up of different materials. So this is called composite wire.
The length of each wire is \[l\] and their area of cross section is \[A\] . The wires are connected in series. The young’s modulus of wire is given by Y and 2Y. Now applying force \[F\]across the wire. Due to the force applied, the wire will be stretched. So we need to calculate the elongation produced on wire due to the force applied.
Let us consider elongation in first wire of young’s modulus Y is \[\Delta {{l}_{1}}\] and that of 2Y is \[\Delta {{l}_{2}}\]
Let the total elongation be \[\Delta l\]
Therefore \[\Delta l=\Delta {{l}_{1}}+\Delta {{l}_{2}}\]….(1)
We know young’ s modulus is ratio of stress to strain
\[Y=\dfrac{{}^{F}/{}_{A}}{{}^{\Delta l}/{}_{l}}=\dfrac{Fl}{A\Delta l}\]
Substituting the values from equation (1)
\[\Delta l=\dfrac{Fl}{AY}\]
The force across both the wires will be the same. Therefore, tension in both the wires will be the same because it is a composite.
For first wire \[\Delta {{l}_{1}}=\dfrac{Fl}{AY}\]
Similarly for second wire \[\Delta {{l}_{2}}=\dfrac{Fl}{A(2Y)}\]
Substituting value in equation (1)
\[\begin{align}
& \Delta l=\dfrac{Fl}{AY}+\dfrac{Fl}{A(2Y)} \\
& \Delta l=\dfrac{3}{2}\dfrac{Fl}{AY} \\
\end{align}\]
Thus on applying force, the total elongation on wire will be \[\dfrac{3}{2}\dfrac{Fl}{AY}\].
Hence option D is correct.
Note:
Young’s modulus describes the physical change in a solid. It is a mechanical property of any solid. Young’s modulus depends on the force applied on the solid and the change in its length. Young’s modulus is different for different materials.
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